Question

2 x square minus 5 upon 6 x + 1 upon 12

Answer 1

Answer:

$\implies 2x^2 - \dfrac{5}{6} x+\dfrac{1}{12}\\\\\\\implies \dfrac{24x^2 - 5(2)x +1}{12}\\\\\\\implies \dfrac{1}{12}\bigg\{ 24x^2 - 10x + 1\bigg\}\\\\\\\implies\dfrac{1}{12}\bigg\{24x^2 - (6+4)x+1\bigg\}\\\\\\\implies\dfrac{1}{12}\bigg\{24x^2-6x-4x+1\bigg\}\\\\\\\implies\dfrac{1}{12}\bigg\{6x(4x-1)-(4x-1)\bigg\}$

= > (1/12) ( 4x - 1 )( 6x - 1 )

or,

(4x-1)(6x-1)/12

Answer 2

Step-by-step explanation:

$\blue{\bold{\underline{\underline{Given/ quadric /equation}}}}$

$2 {x}^{2} - \frac{5x}{6} + \frac{1}{12}$

LCM of 6 and 12 is 12

$= \frac{24 {x}^{2} - 10x + 1}{12}$

$\blue{\bold{\underline{Splitting/ the/ middle/ term}}}$

$\implies </p><p>[tex]( \frac{1}{12} )(24 {x}^{2} - 6x - 4x + 1)$

$= ( \frac{1}{12} )(6x(4x - 1) - 1(4x - 1))$

$= ( \frac{1}{12} )((4x - 1)(6x - 1))$

Therefore,

$2 {x}^{2} - \frac{5x}{6} + \frac{1}{12} = ( \fr</u></em></strong><strong><em><u>d</u></em></strong><strong><em><u>ac{1}{12} )((4x - 1)(6x - 1))$